Problem: You are given three integers a, b, k.

Find two binary integers x and y (x ≥ y) such that 

  1. both x and y consist of a zeroes and b ones; 
  2. x - y (also written in binary form) has exactly k ones. 

You are not allowed to use leading zeros for x and y. 

Input

The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.

Output

If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.

Otherwise print "No".

If there are multiple possible answers, print any of them.

Examples

Input


4 2 3


Output


Yes
101000
100001


Input


3 2 1


Output


Yes
10100
10010


Input


3 2 5


Output


No

Note

In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.

In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.

In the third example, one may show, that it's impossible to find an answer.

Solution (in CPP):

Solution:
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
template<typename T> using PQ = priority_queue<T>;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define lb(c, x) distance((c).begin(), lower_bound(all(c), (x)))
#define ub(c, x) distance((c).begin(), upper_bound(all(c), (x)))
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,a,n) for(ll i=a;i<(ll)(n);++i)
#define repr(i,a,n) for(ll i=n-1;(ll)a<=i;--i)
#define ENDL '\n'
typedef pair<int,int> Pi;
typedef pair<ll,ll> PL;
constexpr ll mod = 1000000007; // 998244353;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 +99);
const ld PI = acos((ld)-1);
constexpr ll dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
template<typename T,typename U> inline bool chmin(T& t, const U& u){if(t>u){t=u;return 1;}return 0;}
template<typename T,typename U> inline bool chmax(T& t, const U& u){if(t<u){t=u;return 1;}return 0;}
template<typename T> inline T gcd(T a,T b){return b?gcd(b,a%b):a;}
inline void Yes() { cout << "Yes" << ENDL; }
inline void No() { cout << "No" << ENDL; }
inline void YES() { cout << "YES" << ENDL; }
inline void NO() { cout << "NO" << ENDL; }

template<typename T,typename Y> inline T mpow(T a, Y n) {
  T res = 1;
  for(;n;n>>=1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}

template <typename T>
vector<T> finddivisor(T x) {  //整数xの約数(xを含む)
  vector<T> divisor;
  for(T i = 1; (i * i) <= x; i++) {
    if(x % i == 0) {
      divisor.push_back(i);
      if(i * i != x) { divisor.push_back(x / i);}
    }
  }
  sort(divisor.begin(), divisor.end());
  return divisor;
}

template <typename T> V<T> prefix_sum(const V<T>& v) {
  int n = v.size();
  V<T> ret(n + 1);
  rep(i, 0, n) ret[i + 1] = ret[i] + v[i];
  return ret;
}

template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(auto&& x: vec) is >> x;
  return is;
}

template<typename T,typename Y>
ostream& operator<<(ostream& os,const pair<T,Y>& p){
  return os<<"{"<<p.fs<<","<<p.sc<<"}";
}

template<typename T> ostream& operator<<(ostream& os,const V<T>& v){
  os<<"{";
  for(auto e:v)os<<e<<",";
  return os<<"}";
}

template<typename ...Args>
void debug(Args&... args){
  for(auto const& x:{args...}){
    cerr<<x<<' ';
  }
  cerr<<ENDL;
}



signed main(){
  cin.tie(0);cerr.tie(0);ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(20);
  ll a,b,k;cin>>a>>b>>k;

  if(k==0){
    Yes();
    rep(i,0,b)cout<<1;
    rep(i,0,a)cout<<0;
    cout<<ENDL;
    rep(i,0,b)cout<<1;
    rep(i,0,a)cout<<0;
    cout<<ENDL;
    return 0;
  }

  if(a+b-1<=k || a==0 || b==1)No();
  else{
    V<int> x(a+b,0),y=x,z=x;
    y[0]=1;
    repr(i,a+1,a+b){
      y[i]=1;
    }

    ll rem=k;
    rep(i,a+1,min(a+1+k,a+b)){
      z[i]=1;
      --rem;
    }
    repr(i,2,a+1){
      if(rem==0)break;
      z[i]=1;
      --rem;
    }

    int f=0;
    repr(i,0,a+b){
      x[i]=(f+y[i]+z[i])%2;
      if(2<=f+y[i]+z[i])f=1;
      else f=0;
    }
    if(f==1){
      No();
      return 0;
    }
    //debug(x,y,z);

    Yes();
    for(auto&& i:x)cout<<i;
    cout<<ENDL;
    for(auto&& i:y)cout<<i;
    cout<<ENDL;

    sort(all(x));
    sort(all(y));
  }

}
//(・_・)(・_・)(・_・)(・_・)
//CHECK overflow,what to output?
//any other simpler approach?